Edexcel Jan 2017 (IAL) Paper 5 Q24 (with explained solutions) |

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Edexcel Jan 2017 (IAL) Paper 5 Q24

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SECTION C Answer ALL the questions. Write your answers in the spaces provided. 24 The +6 oxidation state There are many compounds that contain an element in the +6 oxidation state, when six of the elements electrons are used in bonding. Similarities exist between sulfur(VI) compounds and chromium(VI) compounds. 2, while For example, sulfur forms sulfate(VI) ions, SO4 2. The ions chromium forms chromate(VI) ions, CrO4 with corresponding formulae have the same shape. 2, and disulfate(VI) ions, S2O7 2, and dichromate(VI) ions, Cr2O7 Other transition metal compounds exist with the metal in oxidation state +6. Potassium manganate(VI), K2MnO4, can be prepared in alkaline solution but it readily disproportionates in neutral or acidic solution. Potassium ferrate(VI), K2FeO4, is obtained by heating iron filings with potassium nitrate and then pouring the cooled mixture into water. 2. (a) (i) Draw a dot and cross diagram for a sulfate(VI) ion, SO4 2. (ii) State the shape of a sulfate(VI) ion, SO4 (2) (1)26 *P48373A02632* D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A<br />
(iii) Many sulfate(VI) compounds are soluble in water. However, lead(II) sulfate(VI) is sparingly soluble. 0.0500 g of lead(II) sulfate(VI), PbSO4, was mixed with 250.0 cm3 of water. Calculate the mass of undissolved lead(II) sulfate(VI) in this mixture. Data: molar mass of PbSO4 = 303.3 g mol1 solubility of lead(II) sulfate(VI) = 1.26 x 104 mol dm3 (b) (i) Sodium disulfate(VI) decomposes on heating to form only two products: sodium sulfate(VI) and an oxide of sulfur in oxidation state +6. Write the equation for this decomposition. State symbols are not required. (2) (1) (ii) Ammonium dichromate(VI) decomposes on heating to form chromium(III) oxide, nitrogen and water. Write the equation for this decomposition. State symbols are not required. (1) A E R A S I H T N I E T I R W T O N O D A E R A S I H T N I E T I R W T O N O D A E R A S I H T N I E T I R W T O N O D *P48373A02732* 27 Turn over<br />
(c) Use these standard electrode potentials to answer the following questions. Electrode reaction SO4 2(aq) + H2O(l) + 2e (cid:85)(cid:3)(cid:3)SO3 2(aq) + 2OH(aq) CrO4 2(aq) + MnO4 (aq) + e (cid:85)(cid:3)(cid:3)MnO4 2(aq) MnO4 2(aq) + 2H2O(l) + 2e (cid:85)(cid:3)(cid:3)MnO2(s) + 4OH(aq) 2H+(aq) + O2(g) + 2e (cid:85)(cid:3)(cid:3)H2O2(aq) Cr2O7 2(aq) + 14H+(aq) + 6e (cid:85) 2Cr3+(aq) + 7H2O(l) MnO4 (aq) + 8H+(aq) + 5e (cid:85) Mn2+(aq) + 4H2O(l) H2O2(aq) + 2H+(aq) + 2e (cid:85)(cid:3)(cid:3)2H2O(l) FeO4 2(aq) + 8H+(aq) + 3e (cid:85)(cid:3)(cid:3)Fe3+ + 4H2O(l) E (cid:57)/V 0.93 +0.56 +0.59 +0.68 +1.33 +1.51 +1.77 +2.20 (i) Complete the table using information from page 14 of the Data Booklet. (ii) Give the formula of the species which, under standard conditions, is the strongest oxidising agent. (1) (1)(iii) Write the equation for the disproportionation of the manganate(VI) ion. Include state symbols. (1) 28 *P48373A02832* D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A<br />
(iv) Calculate the standard cell potential, E (cid:57) cell, for the following reaction. 2MnO4 (aq) + 6H+(aq) + 5H2O2(aq) (cid:111) 2Mn2+(aq) + 5O2(g) + 8H2O(l) (1) A E R A S I H T N I E T I R W T O N O D (v) Draw a fully labelled diagram of the apparatus you would use to determine the standard electrode potential of the dichromate(VI) / chromium(III) electrode at 298 K, using a standard hydrogen electrode. (5) A E R A S I H T N I E T I R W T O N O D A E R A S I H T N I E T I R W T O N O D *P48373A02932* 29 Turn over<br />
(d) The mass of ethanol in 5.00 cm3 of white wine is found by oxidising the ethanol to ethanoic acid using acidified potassium dichromate(VI) solution. The excess acidified potassium dichromate(VI) solution is then determined. Step 1 Step 2 5.00 cm3 of white wine was diluted to 100.0 cm3 with distilled water. 10.0 cm3 of acidified potassium dichromate(VI) solution, of concentration 0.0150 mol dm3, was placed in a conical flask with 1.00 cm3 of the diluted white wine and left until all the ethanol had been completely oxidised. 10 cm3 (an excess) of potassium iodide solution was added to the flask to react with the remaining potassium dichromate(VI) solution. The iodine produced was titrated with 0.0300 mol dm3 sodium thiosulfate solution. The procedure was repeated twice more. Step 3 Step 4 Step 5 The mean titre of sodium thiosulfate solution was 9.20 cm3. The equations for the reactions taking place are 2Cr2O7 Cr2O7 2S2O3 2 + 16H+ + 3C2H5OH (cid:111) 4Cr3+ + 11H2O + 3CH3COOH 2 + 14H+ + 6I (cid:111) 2Cr3+ + 3I2 + 7H2O 2 + I2 (cid:111) S4O6 2 + 2I (i) Use the equations to determine the mole ratio of S2O3 2 to Cr2O7 2 in this series of reactions. (1) D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A 30 *P48373A03032*<br />
A E R A S I H T N I E T I R W T O N O D A E R A S I H T N I E T I R W T O N O D A E R A S I H T N I E T I R W T O N O D *(ii) Calculate the mass of ethanol in 5.00 cm3 of the original white wine. (6) (Total for Question 24 = 23 marks) TOTAL FOR SECTION C = 23 MARKS TOTAL FOR PAPER = 90 MARKS *P48373A03132* 31<br />

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Section C Question Number 24(a)(i) Acceptable Answers Reject Mark ALLOW any combination of dots and crosses and just dots or just crosses ALLOW any other symbol for extra electrons eg * ALLOW overlapping circles with electrons in correct places IGNORE missing brackets and charge 2 double bonds and 2 single bonds (1) Rest of diagram correct Conditional on M1 (1) IGNORE other diagrams, such as displayed formula IGNORE shape ALLOW 4 single bonds between S and O (1) 4 double bonds Rest of diagram correct Conditional on M1 (1) Acceptable Answers Reject Mark Question Number 24(a)(ii) Tetrahedral ALLOW triangular based pyramidal IGNORE pyramidal WCH0501 1701 Results<br />
Reject Mark Question Number 24(a)(iii) Correct answer with no working scores (2) marks Acceptable Answers EITHER mass of PbSO4 dissolved in 250.0 cm3 = 1.26 x 10-4 x 303.3 x 250.0 1000 = 9.55395 x 10-3 (g) (1) mass undissolved PbSO4 = 0.0500 - 9.55395 x 10-3 = 0.040446 / 0.04045/ 0.0404 / 0.04(0) (g) (1) TE on mass dissolved in 250.0 cm3 OR mol PbSO4 added = 0.0500 = 1.6485 x 104 303.3 and mol PbSO4 dissolved in 250 cm3 = 1.26 x 10-4 x 250.0 1000 = 3.15 x 10-5 (1) mol undissolved PbSO4 = 1.6485 x 1043.15 x 10-5 = 1.3335 x 104 and mass undissolved PbSO4 = 1.3335 x 104 x 303.3 = 0.040446 / 0.04045 / 0.0404 / 0.04(0) (g) (1) TE on mol dissolved in 250 cm3 Question Number 24(b)(i) Question Number 24(b)(ii) Acceptable Answers Reject Mark Na2S2O7Na2SO4 + SO3 ALLOW multiples IGNORE state symbols, even if incorrect Acceptable Answers (NH4)2Cr2O7Cr2O3 + N2 + 4H2O ALLOW multiples IGNORE state symbols, even if incorrect Reject Mark WCH0501 1701 Results<br />
Question Number 24(c)(i) Question Number 24(c)(ii) Acceptable Answers Reject Mark Half-cell 2-(aq) + 4H2O(l) + 3e(-)CrO4 Cr(OH)3(s) + 5OH-(aq) and 0.13 (V) ALLOWIGNORE missing state symbols IGNORE square brackets around Cr(OH)3 Acceptable Answers Reject Mark 2-(aq) / FeO4 2- FeO4 Additional species Question Number 24(c)(iii) 3MnO4 2-(aq) + 2H2O(l)2MnO4 State symbols are required Acceptable Answers Reject Mark -(aq) + MnO2(s) + 4OH-(aq) Question Number 24(c)(iv) Acceptable Answers Reject Mark (+)0.83 (V) / .83 (V) 0.83 (V) WCH0501 1701 Results<br />
Question Number 24(c)(v) Acceptable Answers Mark hydrogen gas 1 atm platinum 1 mol dm3 H+(aq) salt bridge platinum 2 and 1 mol dm3 w.r.t. Cr2O7 Cr3+ acidified/ H+(aq) ALLOW half-cells reversed Hydrogen half-cell: Solution 1 mol dm-3 H+(aq) and platinum (black) electrode ALLOW 1 mol dm-3 hydrochloric acid / HCl / nitric acid / HNO3 ALLOW 0.5 mol dm3 sulfuric acid (1) Hydrogen gas at 1 atm / 1.01 x 105 Pa / 100 kPa pressure / 1 bar (1) Chromium half-cell: Solution 1 mol dm-3 / equimolar with respect to dichromate / Cr2O7 beaker) (1) Acidified / H+(aq) / HCl IGNORE concentration of acid and platinum electrode (1) Connections: Salt bridge dipping into both solutions and voltmeter to measure Standard Electrode Potential and complete circuit ALLOW a salt bridge drawn and just labelled with the electrolyte Do not award this mark if the circuit is incorrect, e.g a cell is included. Ignore ammeter. (1) 2- ions and chromium(III) / Cr3+ ions (in the same WCH0501 1701 Results<br />
Question Number 24(d)(i) Acceptable Answers Reject Mark 6 : 1 OR 6 1 ALLOW 2 = 1 : 6 / 1 Cr2O7 6 IGNORE all working 2 : S2O3 WCH0501 1701 Results<br />
Reject Mark Question Number *24(d)(ii) Correct answer with no working scores (6) Acceptable Answers 2 = 0.030 x 9.20/1000 2 = 0.015 x 10.0/1000 2 reacted with C2H5OH 2 left = 2.76 x 10-4/6 = 4.60 x 10-5 2 left = 1.38 x 10-4/3 = 4.60 x 10-5 M1 mol S2O3 = 2.76 x 10-4 (1) M2 mol Cr2O7 OR mol I2 = 2.76 x 10-4/2 = 1.38 x 10-4 and mol Cr2O7 TE on mol ratio in (i) (1) M3 original mol Cr2O7 = 1.50 x 10-4 (1) M4 mol Cr2O7 = 1.50 x 10-44.60 x 10-5 = 1.04 x 10-4 and mol C2H5OH in 1.00 cm3 diluted wine = 1.04 x 10-4 x 3/2 = 1.56 x 10-4 TE on original mol Cr2O7 C2H5OH (1) M5 mol C2H5OH in 100 cm3 diluted wine / 5.00 cm3 original wine = 1.56 x 10-4 x 100 = 1.56 x 10-2 TE on mol C2H5OH in 1.00 cm3 diluted wine (1) M6 mass C2H5OH in 5.00 cm3 original wine = 1.56 x 10-2 x 46 = 0.7176 / 0.718 / 0.72 (g) TE on mol C2H5OH in 100 cm3 diluted wine/5.00 cm3 original wine (1) IGNORE SF except 1 SF 2 and mol Cr2O7 2 reacted with (Total for Question 24 = 23 marks) WCH0501 1701 Results<br />

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